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MATLAB EXAMPLES Initial-value problems 58:111 Numerical Calculations Department of Mechanical and Industrial Engineering.

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Presentation on theme: "MATLAB EXAMPLES Initial-value problems 58:111 Numerical Calculations Department of Mechanical and Industrial Engineering."— Presentation transcript:

1 MATLAB EXAMPLES Initial-value problems 58:111 Numerical Calculations Department of Mechanical and Industrial Engineering

2 Some useful functions ode23An explicit one-step Runge-Kutta low order(2 nd -3 rd )solver ode23sAn implicit one-step modified Rosenbrock solver of order two ode23tAn implicit one-step trapezoidal rule using a free interpolate ode23tb An implicit trapezoidal rule followed by a backward differentiation rule of order two ode45An explicit one-step Runge Kutta medium-order (4 th -5 th ) solver ode113 A multistep Adams-Bashforth-Moulton PECE solver of varying order (1 st –13 th ) ode15s An implicit, multistep numerical differentiation solver of varying order (1 st -5 th )

3 IVP The initial value problems for 1 st order differential equation are defined by the following standard form: Example: The exact solution is: In the following slides, we will try to solve this problem by Euler and Runge-Kutta methods in MATLAB. The ODE function was defined as M-file (ex1_odefun.m): % IVP function for EX1 function xdot=ex1_odefun(t,x) xdot=x+1;

4 IVP –Euler method Use Euler method to solve the IVP example, step size sets to 0.01. Write the M-file as follows (ivp-euler.ex1.m): % IVP solution - Euler method % example 1: x'=x+1, x(0)=0, solve x(1) % the range of t [a,b] a=0.0; b=1.0; % the initial value of x x=0.0; %the number of steps n=100; %set the matrix x0=zeros(n,1); % set the step size h=(b-a)/100; %set the initial value of t t=a; for k=1:n % Record the values of x in each step x=x+h*ex1_odefun(t,x); x0(k)=x; t=t+h; end %the final result of x x0(100) Run this M-file: >> ivp_euler_ex1 ans = 1.7048

5 IVP –Runge-Kutta method To use Runge-Kutta method, here we only try ode23 and ode45 (step size =0.1): >> tspan=linspace(0,1,11); >> x0=[0;0]; >> [t,x]=ode23(@ex1_odefun,tspan,x0); >> x(11) ans = 1.7182 >> [t,x]=ode45(@ex1_odefun,tspan,x0); >> x(11) ans = 1.7183 The results show the higher order solver ode45 got more accurate result than ode23 and Euler method.

6 IVP - Plot To show the accuracy of different methods, we need plot the solution together, the following M- file is for plotting the results of exact, Euler and 4 th Runge-Kutta methods (ivp-plot_ex1.m): % IVP solution - plot % example 1 x'=x+1 % the range of t [a,b] a=0.0; b=1.0; % the initial value of x x=0.0; %the number of steps n=10; %set the matrix x1=zeros(n,1); %the step size h=(b-a)/10; %set the initial value of t t=a; %Euler method - Iterations for k=1:n+1 % Record the values of x in each step x1(k)=x; x=x+h*ex1_odefun(t,x); t=t+h; end %plot the results t=linspace(0,1,11); %Runge-Kutta method rx0=[0;0]; [t,x2]=ode45(@ex1_odefun,t,rx0); %exact solution x0=exp(t)-1; plot(t,x0,t,x1,'.',t,x2,'o') xlabel('t') ylabel('x')

7 IVP - Plot h = 0.1h = 0.01 The following figures show the results by different step size h. The solid line is the exact solution, the solid dots represent the result of Euler method, the circles represents the results of 4 th order Runge-Kutta method.

8 Compare the results of different methods (all keep 4 digits after the radix point) in the following table (x(1)): MethodSolution Exact1.7183 Euler (h = 0.1)1.5937 4 th Runge-Kutta (h = 0.1)1.7183 Euler (h = 0.01)1.7084 4 th Runge-Kutta (h = 0.01)1.7183 IVP-Summary


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